Also, read |

Matrix Formulas |

Mean, Median,Mode Formulas |

Set Theory |

# Determinant formulas:

In the previous section we discussed matrices. Now you are much aware of matrices its properties, addition, subtraction and multiplication. Now another term is there which enhances the properties of square matrices. It has a wide range of applications in algebraic equations. As we know that we can express an algebraic equation in the form of matrices and determinant. Thus we will use **determinant formulas** to solve the same.

let’s suppose equation,

a_{1}x + b_{1}y = c_{1}

a_{2}x + b_{2}y = c_{2}

We can write it as

\(\begin{bmatrix} a_{1} & b_{1}\\ a_{2} & b_{1}\\ \end{bmatrix}\) \(\begin{bmatrix} x \\ y\\ \end{bmatrix}\) = \(\begin{bmatrix} c_{1} \\ c_{2}\\ \end{bmatrix}\)

we know that number a_{1 }b_{2} – a_{2} b_{1} which determines either the solution is unique or not is said to be **d****eterminants.**

**Calculation of determinant and evaluating determinant formulas:**

- Determinant of a matrix of 1
^{st}order i.e. (1 x 1) A = [a_{11}]_{1×1}is |a_{11}| = a_{11} - Determinant of for 2 x 2 matrix i.e

Δ = \(|A| = \left| \begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} \right|\)

| A | = a_{11} × a_{22} – a_{21} × a_{21}

- Determinant of for 3 x 3 matrix i.e

Δ = |A| = \(\left| \begin{matrix} a_{11} & a_{12} & a_{13}\cr a_{21} & a_{22} & a_{23} \cr a_{31} & a_{32} & a_{33} \end{matrix}\right|\)

a_{11}\(\left| \begin{matrix} a_{22} & a_{23}\cr a_{32} & a_{33} \end{matrix}\right|\) – a_{12}\(\left| \begin{matrix} a_{21} & a_{23}\cr a_{31} & a_{33} \end{matrix} \right|\) + a_{13}\(\left| \begin{matrix} a_{21} & a_{22}\cr a_{31} & a_{32} \end{matrix} \right|\)

**Properties of determinant:**

- If rows and columns of determinants are interchanged, the value of the determinant remains unchanged.

- From above property, we can say that if A is a square matrix, then det (A) = det (A′), where A′ = transpose of A.

- If any two rows (or columns) of a determinant are interchanged, then sign of determinant changes.
- Also, If any two rows (or columns) of a determinant are identical (i.e all corresponding elements are the same), then the value of determinant is zero.
- If each element of a row (or a column) of determinants are multiplied by a constant k, then its value gets multiplied by k.

- According to this property, we can take out any common factor from any one row or any one column of a given determinant.
- If corresponding elements of any two rows (or columns) of a determinant are proportional (in the same ratio), then its value is zero.

- In a determinant, if some or all elements of a row or column are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants.
- If we add elements of one row (or column){also by multiplying same constants to each element of row or column} to corresponding elements of other rows (or column), the value of determinant remains the same,
- There are many uses of
**determinant formulas.**

We can symbolize this operation by

R_{i} → R_{i} + kR_{j } or C_{i} → C_{i} + k C_{j}

- If Δ
_{1}is the determinant obtained by applying R_{i}→ kR_{i}or C_{i}→ kC_{i}to the determinant Δ, then Δ_{1}= kΔ. - If you carry more than one operation like R
_{i}→ R_{i}+ kR_{j}is done in one step, take care that each operation has its remark so one operation must not be affected by other.

**Area of triangle using determinant formulas:**

By the coordinate formula of area of triangle whose vertices are (x_{1} , y_{1} ), (x_{2} , y_{2}) and (x_{3} , y_{3}) is:

\(\frac{1}{2}\) [x_{1} (y_{2} –y_{3}) + x_{2} (y_{3} –y_{1}) + x_{3} (y_{1} –y_{2})]

Now we express it in determinant form as:

\(\alpha =\frac 12 \left| \begin{matrix} x_1 & y_1 & 1\cr x_2 & y_2 & 1 \cr x_3 & y_3 & 1 \cr \end{matrix} \right|\)

Properties of area of triangles obtained by **determinant formulas:**

- Since we know that area is always a positive quantity, we always take the absolute value of the determinant.
- Another feature of this area is, if area is already there, we can use both positive and negative values of the determinant for calculation.
- Also, the area of the triangle formed by three collinear points is zero.
- Now we will read another section of
**determinant formulas**i.e. minors and cofactors.

**Minors and cofactors:**

**Minor:**The determinant obtained by deleting its i^{th}row and j^{th}column in which element a_{ij}lies is minor of an element a_{ij}of a determinant.

We denote minor of an element a_{ij}by M_{ij}.

- Always remember that minor of an element of a determinant is a determinant of order n – 1. When the actual determinant is of order n where (n ≥ 2).

**Cofactor:**We denote cofactor of an element a_{ij}, by A_{ij}and define it by A_{ij}= (–1)i + j M_{ij}, where M_{ij}is minor of a_{ij}.**Determinant formulas**can be used in a specific way to solve the problems.

**Adjoint and Inverse of a Matrix:**

**Adjoint of a matrix:**We define the adjoint of a square matrix A as the transpose of the cofactor matrix of matrix A.

Actually, we denote adjoint of a matrix by “adj A”

- If A be any given square matrix of order n, then A(adj A) = (adj A) A = |A|I, where I is the identity matrix of order n.

**Singular and non-singular matrix:**

**Singular matrix:**We can say a square matrix A to be singular if |A| = 0.**Non-singular matrix:**We can say a square matrix A to be non-singular if |A|≠ 0

**Properties:**

- One thing to remember is that if A and B are nonsingular matrices of the same order, then AB and BA are also nonsingular matrices of the same order.
- |AB| = |A||B|, where A and B are square matrices of the same order.

**Inverse of a matrix:**

We can say a square matrix A is invertible if and only if A is nonsingular matrix.

If B is square matrix and AB = BA = I, then we describe B as inverse of A.

Also A^{–1} = B or B^{–1} = A and hence (A^{–1})^{–1} = A.

The actual definition of inverse of a matrix is

A^{-1} = \(\frac{1}{|a|}\) *(adj A)*

**Determinant formulas based examples:**

\(\left| \begin{matrix} 102 & 18 & 36\cr 1 & 3 & 4 \cr 17 & 3 & 6 \cr \end{matrix} \right|\)*Evaluate*

**Solution:**As we see that

\(\left| \begin{matrix} 102 & 18 & 36\cr 1 & 3 & 4 \cr 17 & 3 & 6 \cr \end{matrix} \right|\) = \(\left| \begin{matrix} 6(17) & 6(3) & 6(6)\cr 1 & 3 & 4 \cr 17 & 3 & 6 \cr \end{matrix} \right|\) = 6 \(\left| \begin{matrix} 17 & 3 & 6\cr 1 & 3 & 4 \cr 17 & 3 & 6 \cr \end{matrix} \right|\) = 0

*Find the area of the triangle whose vertices are (3, 8), (– 4, 2) and (5, 1).*

**Solution:**Δ = \(\frac 12 \left| \begin{matrix} 3 & 8 & 1\cr -4 & 2 & 1 \cr 5 & 1 & 1 \cr \end{matrix} \right|\)

= \(\frac{1}{2}\) [3(2 – 1) –8(-4 – 5)+1(-4 – 10)]

= \(\frac{1}{2}\) [ 3 + 72 + – 14] = \(\frac{61}{2}\)

## Be First to Comment